# What's the meaning of factors 'collapsing' in quotients like $\mathbb{Z}_4\times\mathbb{Z}_6/\langle(0,1)\rangle$? [duplicate]

My book says that in

$$\mathbb{Z}_4\times\mathbb{Z}_6/\langle(0,1)\rangle$$

since we're setting everything in $\langle(0,1)\rangle$ to $0$, it's like. That is, the whole second factor $\mathbb{Z}_6$ of $\mathbb{Z}_4\times\mathbb{Z}_6$ is collapsed, leaving just the factor $\mathbb{Z}_4$.

Then, for

$$\mathbb{Z}_4\times\mathbb{Z}_6/\langle(0,2)\rangle$$

It says that the same thing happens, but now $\mathbb{Z}_6$ is collapsed by a subgroup of order $3$, therefore givinf a group in the second factor of order $2$, so it's isomorphic to $\mathbb{Z}_4\times\mathbb{Z}_2$ I tried to visualize the cosets for both groups, here they are:

$$\mathbb{Z}_4\times\mathbb{Z}_6/\langle(0,1)\rangle$$

$$H = \{(0,0),(0,1),(0,2),(0,3),(0,4),(0,5)\}$$ $$H_1 = \{(1,0),(1,1),(1,2),(1,3),(1,4),(1,5)\}$$ $$H_2 = \{(2,0),(2,1),(2,2),(2,3),(2,4),(2,5)\}$$ $$H_3 = \{(3,0),(3,1),(3,2),(3,3),(3,4),(3,5)\}$$

And

$$\mathbb{Z}_4\times\mathbb{Z}_6/\langle(0,2)\rangle$$

$$H = \{(0,0),(0,2),(0,4)\}$$ $$H_1 = \{(1,0),(1,2),(1,4)\}$$ $$H_2 = \{(2,0),(2,2),(2,4)\}$$ $$H_3 = \{(3,0),(3,2),(3,4)\}$$

But I cannot visualize this 'collapse' thing

## marked as duplicate by Dietrich Burde abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 28 '17 at 14:57

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• quotients make some things equal, so some people like to think of it as collapsing those things together.'' But it's algebra, so you are probably better off not trying to force some intuition and just focus on understanding the rules. – AnonymousCoward Apr 28 '17 at 12:13
• A good way to think about quotient is to consider the quotient group $\mathbb{Z}/2\mathbb{Z}$. This quotient group has 2 elements, the even numbers and the odd numbers. So we "collapse" the even numbers and the odd numbers into a single point respectively. – Alex Vong Apr 28 '17 at 12:17
• The interesting thing is that the quotient group has a natural group structure. For $\mathbb{Z}/2\mathbb{Z}$, it is what we learned in elementary school, even number plus even number is even, odd number plus odd number is even and odd number plus even number is odd. From this, one can easily observe that $\mathbb{Z}/2\mathbb{Z} \cong \mathbb{Z}_2$ – Alex Vong Apr 28 '17 at 12:21

## 1 Answer

Personally I prefer writing $\langle (0,2)\rangle$ as $0\times 2\Bbb Z_6$, looks neater and gives a more intuitive sense when comparing to the traditional ones of integers. Anyway onto your question, collapsing is more of a metaphore with the illustration that it originally stands as a big building that falls into pockets where the material must be.

Personally I just view it as a word meaning that "we group these elements together and declare them to be equal that is compatible with our algebraic structure". A natural question then is when we do this, how does the structure change? What are the differenses? What are the new things? Etc. This leads to us wanting to find isomorphisms to what is viewed as more simple structures that are isomorphic.

Here it is easy to consider things as we have $$\frac{\Bbb Z_4\times\Bbb Z_6}{0\times 2\Bbb Z_6}\cong \frac{\Bbb Z_4}{0}\times\frac{\Bbb Z_6}{2\Bbb Z_6}$$ and the latter one is easy to see is isomorphic to $\Bbb Z_3$ after some work.